Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
int main() {

int N;
cin >> N;
vector<double> dp(N+1);
dp[0] = 1;

for(int coin=0; coin<N; coin++) {
   double p_head;
   cin >> p_head;
   for(int i=coin+1; i>=0; i--) {
      dp[i] = ( i==0 ? 0 : dp[i-1]*p_head ) + dp[i] * (1-p_head);
   }
}
double answer = 0;
for(int i=0; i<=N; i++) {
	if(i>N-i) {
		answer+=dp[i];
	}
}
cout<< fixed << setprecision(10) << answer << "\n";
return 0;
}